TITLE: If a finite group $G$ is solvable, is $[G,G]$ nilpotent?
QUESTION [15 upvotes]: If $G$ is a connected solvable Lie-group, then $[G,G]$ is nilpotent. The corresponding statement for Lie algebras follows from Lie's theorem, and it then follows from connected Lie groups by exponentiation. 
Is the statement also true for finite groups? I can't find a counterexample, but I didn't try that hard.
Motivation: I'm prepping to teach a group theory and Galois theory course, so I'm brainstorming interesting questions about finite groups. Whenever I come up with one I can't answer, I plan to toss it up here or at MO.

REPLY [4 votes]: I just want to throw in that there is actually a term for groups with nilpotent derived subgroups; they are called nilpotent-by-abelian.  An equivalent condition is that $G'\leqslant \text{Fit}(G)$, which isn't always true in a solvable group.
In particular all solvable groups with $\ell_F(G)>2$ are counterexamples, which is another easy way to see that the property does not hold for $S_4$ (which is a $2$-Frobenius group, and thus has Fitting length $3$).